1

The Best Glide speed is the indicated airspeed that guarantees the maximum glide distance during an engine out scenario. However, the landing straight ahead option after an engine failure may not be the best option especially if there is nothing but houses in front of you after an engine failure. Sometimes turning back towards the field may be the best option.

The Best Glide speed is actually tied to a best Lift/Drag ratio for a specific angle of attack. When in a banked turn, the pilot must increase the angle of attack to avoid a spiral dive. So this implies that the best glide speed would increase during a banked turn in order to maintain the best L/D angle of attack. My question is how would one calculate the Best Glide speed for an aircraft in a 30 degree bank based off the POH listed Best Glide at Gross Weight (I do realize best glide varies with weight).

DeltaLima
  • 83,202
  • 11
  • 272
  • 366
DLH
  • 5,799
  • 3
  • 27
  • 50
  • 1
    This doesn't quite add up. (A) You say "When in a banked turn, the pilot must increase the angle of attack to avoid a spiral dive." When gliding, why must the pilot increase the AoA? Let's say you want to fly at the AoA for Min Sink Rate. Isn't this the same AoA in wings-level flight as in a banked turn? Same for stall AoA? Same for best L/D AoA? The concept of increasing AoA when banked pertains more to flying with a motor and trying to maintain a constant altitude. (ctd) – quiet flyer Mar 18 '21 at 19:21
  • I think your presumptions are incorrect. When turning in an engine out glide you still want to maintain best glide speed, which will correlate to best L/D AOA. However, in order to do so you will need to lower the nose, which will increase sink rate. – Michael Hall Mar 18 '21 at 19:21
  • (ctd) Then you say (B) "So this implies that the best glide speed would increase during a banked turn in order to maintain the best L/D angle of attack." --yes, it must increase according to the square root of the change in wing loading . So (B) is true. But (B) doesn't follow logically from (A), because (A) is incorrect. – quiet flyer Mar 18 '21 at 19:23
  • I think you can still fix this w/o invalidating the single answer posted so far, because the answer only addresses (B), not (A). – quiet flyer Mar 18 '21 at 19:25
  • 1
    Another huge complication is that a 180 degree reversal arguably has nothing to do with best L/D at all. You are in no sense trying to maximize forward distance covered. You want to minimize time spent in the reversal and you also want to minimize the sink rate during the reversal. I seem to recall I've seen analyses suggesting a 45-degree banked turn is the way to go, letting the airspeed bleed off (or even pulling the stick aft to help the airspeed bleed off) to near min sink for that bank angle -- but certainly not lower!-- I'd have to look some more to relocate those analyses-- – quiet flyer Mar 18 '21 at 19:31
  • @quietflyer I would like to see those analysis if you can did them up. Perhaps I am thinking about this all wrong. In the airplane world we are taught to maintain best glide when the engine goes out but I do understand that turns can be tighter when travelling at a slower speed. Of course you will need to be careful not to stall the airplane which occurs at a high airspeed when banked. – DLH Mar 18 '21 at 20:06

2 Answers2

4

During a turn the load factor increases. This increases the apparent weight. Thus, during a turn, the best glide speed is selected according to the weight times the load factor.

The load factor at 30 degrees bank is $$n=\frac{1}{\cos{\theta}} \approx 1.2$$

The best glide speed is thus the best glide speed for a weight that is 1.2 times your actual weight. This will be proportional to the square root of the load factor. $$V_{opt,\theta}= \sqrt{\frac{1}{\cos{\theta}}} V_{opt,0^\circ}$$

DeltaLima
  • 83,202
  • 11
  • 272
  • 366
  • Is this really correct? The load factor is only equal to 1/cos(theta) in level flight where you pull on the yoke to maintain altitude. But presumably one will not be attempting to maintain level flight during engine out. – SMeznaric Jun 06 '23 at 15:43
  • @SMeznaric you are right about the level flight assumption being likely not valid during engine out conditions. However at small flight path angles, the formula approximates the load factor well. The correct form (assuming constant flight path angle or gamma) is cos(gamma)/cos(theta) – DeltaLima Jun 06 '23 at 18:04
  • I am not sure that's correct @DeltaLina. If you're not pulling on the yoke and accept a higher descent rate during the turn then your load factor should stay unchanged. – SMeznaric Jun 13 '23 at 16:55
  • @SMeznaric I am note sure what you are trying to achieve. assuming coordinated flight, the load factor is directly related to the bank angle and the flight path angle. – DeltaLima Jun 13 '23 at 17:32
  • I can't tell if your final equation is right or not, I am only taking an issue with your statement about the load factor going up in a turn. This is only true in level flight. And the reason it's true in level flight is because you need more lift to counteract gravity in a turn. But if you're not in level flight, you're trimmed out and turning without pulling, you'll increase your descent rate and then this assumption does not hold. The load factor in a turn in general could in general be higher or lower than the normal level flight load factor. – SMeznaric Jun 14 '23 at 09:52
  • @SMeznaric no, it is not only true in level flight. Also in a flight of constant flight path angle this is true. Of course you can turn briefly with a load factor of 1, but you will accelerate downward during the turn. Very quickly you will develop an very negative vertical speed and thus loose a lot of altitude. Not what you want in an engine out scenario – DeltaLima Jun 14 '23 at 13:40
  • You don't have many options in an engine out. If you're turning you'll have to increase your descent rate or lower your speed. If you lower your speed you're increasing your load factor and getting closer to a stall which in a stressful potentially uncoordinated turn might end up in spinning to the ground. Much preferable to turn without pulling and keep the airplane in the air. – SMeznaric Jun 14 '23 at 14:07
  • I am not stating that you should not increase your descent rate. You will need to do that to maintain speed. Your speed should be as calculated above, and the descent rate adjusted accordingly. The load factor will be above 1, otherwise you will accelerate downward (meaning you will descent at an increasingly higher rate) – DeltaLima Jun 14 '23 at 14:43
  • OK, so since your speed has gone up and your angle of attack is reduced, you accept then that the load factor going up to 1 / cos(theta) statement is not correct then? That's all I'm saying so it seems that we're on the same page. – SMeznaric Jun 14 '23 at 17:13
  • Your speed has gone up, the angle of attack has stayed the same. The load factor has increased by a factor 1/cos(theta). In the transition into the turn, the load factor will be lower, but once established in the turn, cos(gamma)/(cos(theta) is the load factor. – DeltaLima Jun 14 '23 at 18:07
  • If forces don't balance during the turn then your load factor can be anything independently of the angle of bank. But anyway, we're not getting anywhere, so I'll leave it here. – SMeznaric Jun 16 '23 at 12:20
2

In a descending turn, maintain the same glide speed. Your rate of descent will increase, but you can fly the same speed.

The issue of load factor in a banked turn applies to a level turn. In a descent, fly your glide speed and let the descent rate increase.

Will
  • 1,107
  • 4
  • 7